Radiation pressure from light source

  • #1
528
33
The energy of photon is $$E=\frac{hc}{\lambda}$$
Now if we have an isotropic point light source of power P,
Number of photons $$N=\frac{P}{E} = \frac{P \lambda}{hc}$$
Hence one can find the change in momentum and hence the force exerted by a beam or light sources.
But lets say we keep an isotropic light source at a point on the optical axis of a convex lens such that rays emerge parallel to axis after refraction.The lens does not absorb nor reflect any light energy.
Then the force exerted by the light source on lens is given by:
$$F=\frac{P}{2c}(1-\frac{d}{\sqrt [2]{r^2 + d^2}} - \frac{r^2}{2(r^2 + d^2)})$$
How do derive the expression?
My attempt:
Lets say a ray approaches at an angle ##\theta##
Then initial momentum of a single photon = ##m(vcos\theta + vsin\theta )## and final momentum is ##mv##
Hence, $$\delta p = \frac {h}{\lambda}(1-cos\theta -sin\theta)$$
 

Answers and Replies

  • #2
516
63
The energy of photon is $$E=\frac{hc}{\lambda}$$
Now if we have an isotropic point light source of power P,
Number of photons $$N=\frac{P}{E} = \frac{P \lambda}{hc}$$
Hence one can find the change in momentum and hence the force exerted by a beam or light sources.
But lets say we keep an isotropic light source at a point on the optical axis of a convex lens such that rays emerge parallel to axis after refraction.The lens does not absorb nor reflect any light energy.
Then the force exerted by the light source on lens is given by:
$$F=\frac{P}{2c}(1-\frac{d}{\sqrt [2]{r^2 + d^2}} - \frac{r^2}{2(r^2 + d^2)})$$
How do derive the expression?
My attempt:
Lets say a ray approaches at an angle ##\theta##
Then initial momentum of a single photon = ##m(vcos\theta + vsin\theta )## and final momentum is ##mv##
Hence, $$\delta p = \frac {h}{\lambda}(1-cos\theta -sin\theta)$$
What's m? The mass of a photon is zero, instead the photon's momentum is given by ##p=E/c##
 
  • #3
528
33
What's m? The mass of a photon is zero, instead the photon's momentum is given by ##p=E/c##
That's only for relation. I am relating this to a case where you have a ball instead of photon but the final expression is correct
 
  • #4
Drakkith
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.The lens does not absorb nor reflect any light energy.
Then the force exerted by the light source on lens is given by:
I don't think the light can exert a force if it isn't reflected or absorbed.
 
  • #5
Andy Resnick
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<snip>
But lets say we keep an isotropic light source at a point on the optical axis of a convex lens such that rays emerge parallel to axis after refraction.The lens does not absorb nor reflect any light energy.<snip>
This is more interesting than I thought: on the face of it, there is no net force because the light field appears symmetric: by analogy, objects located in the center of my optical trap experience no net force.

However, there is a the change in momentum along the optical axis, because a ray enters the lens at angle Θ with respect to the optical axis and leaves the lens parallel to the optical axis. In optical trapping it's sometimes referred to as the 'scattering force', and the previous paragraph then refers to the 'gradient force'. The formula you presented likely quantifies that; it looks familiar but I can't recall the context...
 
  • #6
Drakkith
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However, there is a the change in momentum along the optical axis, because a ray enters the lens at angle Θ with respect to the optical axis and leaves the lens parallel to the optical axis. In optical trapping it's sometimes referred to as the 'scattering force', and the previous paragraph then refers to the 'gradient force'.
Interesting. I wouldn't have thought of refracting the light as scattering. I assume that if the light beam changes direction between entering and exiting the lens then it must transfer momentum to the lens?
 
  • #7
A.T.
Science Advisor
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I assume that if the light beam changes direction between entering and exiting the lens then it must transfer momentum to the lens?
Of course, conservation of momentum.
 
  • #8
528
33
This is more interesting than I thought: on the face of it, there is no net force because the light field appears symmetric: by analogy, objects located in the center of my optical trap experience no net force.

However, there is a the change in momentum along the optical axis, because a ray enters the lens at angle Θ with respect to the optical axis and leaves the lens parallel to the optical axis. In optical trapping it's sometimes referred to as the 'scattering force', and the previous paragraph then refers to the 'gradient force'. The formula you presented likely quantifies that; it looks familiar but I can't recall the context...
I have posted a picture of my attempt. I not good in LaTeX code. Please go through it and correct my mistakes if you can. It will benefit me a lot. I am close to the answer
 
  • #10
528
33
I agree!
Where am I going wrong?
These type of question comes in Olympiads and entrance exams.
Why not a separate forum for Olympiads and entrance exams?
 
  • #11
528
33
I have been trying this problem for 2 days.
Where am I going wrong? I have uploaded the working. (Picture oriented sideways and deleted by Mod)
 
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